Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03
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A module homomorphism vanishing on N factors uniquely through M/N

Statement

Let f:M→P be a module homomorphism and let N≤M satisfy N⊆ker⁡f. There is a unique module homomorphism

fˉ:M/N⟶P

such that fˉ(m+N)=f(m), equivalently f=fˉ∘π.

Facts & Assumptions

Given: A module homomorphism f:M→P and a submodule N≤M with N⊆ker⁡f.

[L1]

The canonical map π:M→M/N is a surjective module homomorphism with kernel N (The canonical map M→M/N is a surjective module homomorphism with kernel N; thus every submodule is a kernel).

[L2]

A group homomorphism that kills a normal subgroup factors uniquely through the group quotient (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L3]

A module homomorphism is additive and scalar-preserving, and its kernel is the preimage of zero (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · constructive
1.1

By [L1], the canonical map π:M→M/N is an additive-group quotient map with kernel N. Viewing the modules as additive groups, [L3] makes f a group homomorphism and the hypothesis says it kills N. By [L2], define the unique additive homomorphism fˉ by fˉ(m+N)=f(m), with f=fˉ∘π.

L1L2L3givenconstruct
2.1

For every coset, fˉ(r(m+N))=fˉ(rm+N)=f(rm)=rf(m)=rfˉ(m+N), so fˉ is scalar-preserving.

step 1.1L3given
3.1

Thus fˉ is a module homomorphism with the required factorisation.

step 1.1step 2.1L3
4.1

Any module-homomorphism factor is in particular an additive-group factor, so the uniqueness in step 1.1 proves its uniqueness as a module homomorphism.

step 1.1L2L3discharge-construct: final∎

Depends on

Used by

Dependency tree · two levels

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Sources